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				<span style="position: absolute;left:15px;bottom:15px;width:90%;"><font class="view-text" style="color:#fcfcfc;font-size:25px">拉格朗日反演 学习小记</font><br><a href="/tags/2021/" class="tag"><span  style="background-color: rgb(52, 152, 219);">2021</span></a>&nbsp;<a href="/tags/拉格朗日反演/" class="tag"><span  style="background-color: rgb(231, 76, 60);">拉格朗日反演</span></a>&nbsp;<a href="/tags/笔记/" class="tag"><span  style="background-color: rgb(82, 196, 26);">笔记</span></a></span>
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                <h2 id="_1">拉格朗日反演</h2>
<p><strong>多项式复合</strong>：<script type="math/tex">F(G(x))=\sum\limits_{i=0}f_iG(x)^i</script>
</p>
<p>不过我们这样的形式看着不是很舒服，将 <strong>复合</strong> 看做是二元运算，将 <script type="math/tex">F\circ G</script> 看做是 <script type="math/tex">F(G(x))</script>。</p>
<p>若不特殊说明，以下考虑常数项为 <script type="math/tex">0</script>，一次项系数不为 <script type="math/tex">0</script> 的多项式。</p>
<p>所有满足这个条件的多项式构成一个群，满足以下条件：</p>
<ol>
<li>
<p><strong>封闭性</strong>：两个多项式复合得到的多项式常数为 <script type="math/tex">0</script>，一次项不为 <script type="math/tex">0</script>
</p>
</li>
<li>
<p><strong>结合律</strong>：<script type="math/tex">(F\circ G)\circ H</script> 与 <script type="math/tex">F \circ(G\circ H)</script> 是等价的，可以拆开来得到</p>
</li>
<li>
<p><strong>单位元</strong>：<script type="math/tex">e</script> 就是 <script type="math/tex">x</script>，显然 <script type="math/tex">A\circ e=e\circ A=A</script>
</p>
</li>
<li>
<p><strong>逆元</strong>：每个多项式有得到唯一解的构造，故有可逆性。</p>
</li>
</ol>
<p><strong>复合逆</strong>：有两个多项式 <script type="math/tex">F(x)</script> 和 <script type="math/tex">G(x)</script> 满足 <script type="math/tex">F\circ G=e</script> 则称 <script type="math/tex">F,G</script> 互为复合逆。</p>
<blockquote>
<p>
<script type="math/tex">F\circ G=e</script> 与 <script type="math/tex">G\circ F=e</script> 是等价的。</p>
</blockquote>
<p><strong>结论</strong>：若 <script type="math/tex">G\circ F=e</script>，则：
<script type="math/tex; mode=display">[x^n]G(x)=\frac1n[x^{-1}]F(x)^{-n}</script>
<strong>证明</strong>：显然此时的 <script type="math/tex">F(x)</script> 无法求逆，因为常数项为 <script type="math/tex">0</script>，我们需要引入 <strong>分式域</strong>。</p>
<p>如果有 <script type="math/tex">F(x)</script> 无法求逆，那么必定可以找出整式 <script type="math/tex">G(x)=\dfrac {F(x)}x^k</script> 使得其可以求逆，此时 <script type="math/tex">F(x)</script> 的逆即为 <script type="math/tex">x^{-k}G(x)^{-1}</script>
</p>
<p>
<script type="math/tex">G(x)^{-1}</script> 是整式再乘上 <script type="math/tex">x^{-k}</script> 自然就出现了负数次项。</p>
<blockquote>
<p><strong>引理</strong>：对于满足一开始定义的 <script type="math/tex">F(x)</script>，有:
<script type="math/tex; mode=display">[x^{-1}]F^\prime(x)F(x)^k=[k==-1]</script>
</p>
<ul>
<li>
<p>若 <script type="math/tex">k\neq-1</script>，则 <script type="math/tex">F^\prime(x)F(x)^k=(\frac 1{k+1}F(x)^{k+1})^\prime</script>，是一个整式</p>
</li>
<li>
<p>若 <script type="math/tex">k=-1</script>，则 
<script type="math/tex; mode=display">[x^{-1}]F^\prime(x)/F(x)=[x^{-1}]F^\prime(x)\times x^{-1} \left(\frac{F(x)}x\right)^{-1}=[x^0]F^\prime(x)\left(\frac{F(x)}x\right)^{-1}</script>
只需要考虑 <script type="math/tex">[x^0]F^\prime(x)=[x^1]F(x)</script>，<script type="math/tex">[x^0]\left(\frac{F(x)}x\right)^{-1}=\left([x^1]F(x)\right)^{-1}</script>，乘起来得到 <script type="math/tex">1</script>
</p>
</li>
</ul>
</blockquote>
<p>现在可以开始证哩！
<script type="math/tex; mode=display">
\begin{aligned}
G\circ F&=e\\
\sum_{i=0}g_iF(x)^i&=x\\
(\sum_{i=0}g_iF(x)^i)^\prime&=x^\prime\\
\sum_{i=0}g_iiF(x)^{i-1}F^\prime(x)&=1\\
\sum_{i=0}g_iiF(x)^{i-1-n}F^\prime(x)&=F(x)^{-n}\\
\sum_{i=0}g_ii[x^{-1}]F(x)^{i-1-n}F^\prime(x)&=[x^{-1}]F(x)^{-n}\\
\sum_{i=0}g_ii[i-1-n=-1]&=[x^{-1}]F(x)^{-n}\\
ng_n&=[x^{-1}]F(x)^{-n}\\
[x^n]G(x)&=\frac1n[x^{-1}]F(x)^{-n}
\end{aligned}
</script>
这个 <script type="math/tex">[x^{-1}]</script> 系数还是好麻烦呀QwQ
<script type="math/tex; mode=display">[x^{-1}]F(x)^{-n}=[x^{-1}]x^{-n}\left(\frac {F(x)}x\right)^{-n}=[x^{n-1}]\left(\frac {F(x)}x\right)^{-n}</script>
<strong>扩展</strong>：若 <script type="math/tex">G\circ F=e</script>，<script type="math/tex">H(x)</script> 为任意多项式，有
<script type="math/tex; mode=display">[x^n](H\circ G)(x)=[x^{-1}]H^\prime(x)F(x)^{-n}</script>
证明过程与上面类似，记 <script type="math/tex">T=H\circ G</script>，来愉快地推柿子：
<script type="math/tex; mode=display">\begin{aligned}
    G\circ F&=e\\
    H\circ G\circ F&=H\\
    T\circ F&=H\\
    \sum_{i=0}t_iF(x)^i&=H(x)\\
    (\sum_{i=0}t_iF(x)^i)^\prime&=H^\prime(x)\\
    \sum_{i=0}t_iiF(x)^{i-1}F^\prime(x)&=H^\prime(x)\\
    \sum_{i=0}t_iiF(x)^{i-1-n}F^\prime(x)&=H^\prime(x)F(x)^{-n}\\
    \sum_{i=0}t_ii[x^{-1}]F(x)^{i-1-n}F^\prime(x)&=[x^{-1}]H^\prime(x)F(x)^{-n}\\
    \sum_{i=0}t_ii[i-1-n=-1]&=[x^{-1}]H^\prime(x)F(x)^{-n}\\
    t_n&=\frac1n[x^{-1}]H^\prime(x)F(x)^{-n}\\
    [x^n](H\circ G)(x)&=[x^{-1}]H^\prime(x)F(x)^{-n}
\end{aligned}</script>
</p>
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